剑指 Offer 68 二叉树的最近公共祖先

Posted 2022-05-12 Billy Miracle

剑指 Offer 68 - I. 二叉搜索树的最近公共祖先

题目描述

菜鸡思路：

1. 二叉树，就是递归
2. 二叉搜索树，还可以二分查找避免递归

菜鸡代码：

递归

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) 
 * ;
 */
class Solution 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        if (p->val < root->val && q->val < root->val) 
            root = lowestCommonAncestor(root->left, p, q);
        
        if (p->val > root->val && q->val > root->val) 
            root = lowestCommonAncestor(root->right, p, q);
        
        return root;
    
;

二分查找

class Solution 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        int min = p->val < q->val ? p->val : q->val;
        int max = p->val + q->val - min;
        TreeNode *result = root;
        while (root) 
            if (root->val >= min && root->val <= max) 
                return root;
             else if (root->val > max) 
                root  = root->left;
             else if (root->val < min) 
                root = root->right;
            
        
        return result;
    
;

剑指 Offer 68 - II. 二叉树的最近公共祖先

题目描述

菜鸡思路：

转化为二叉搜索树，利用上一道题的代码

菜鸡代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode 
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) 
 * ;
 */
class Solution 
public:
    unordered_map<TreeNode*, int>myMap;
    int count = 0;
    void search(TreeNode* root) 
        if (!root) 
            return;
        
        search(root->left);
        myMap.insert(root, count++);
        search(root->right);
    
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        search(root);
        int min = myMap[p] < myMap[q] ? myMap[p] : myMap[q];
        int max = myMap[p] + myMap[q] - min;
        TreeNode *result = root;
        while (root) 
            if (myMap[root] >= min && myMap[root] <= max) 
                return root;
             else if (myMap[root] > max) 
                root  = root->left;
             else if (myMap[root] < min) 
                root = root->right;
            
        
        return result;
    
;

大佬思路：

对于一颗树来说，如果节点p、q分别分布在root的左右子树中，那么节点p、q的最近公共祖先为root；如果p、q均只分布在root的一侧（左子树或者是右子树）中，我们则可以递归的处理相同的问题，即在一颗子树（root->left或者root->right）中判断节点p、q的最近公共祖先。

大佬代码：

class Solution 
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) 
        if(!root) 
            return NULL;
        
        if(root == p || root == q) 
            return root;
        
        auto left = lowestCommonAncestor(root->left, p, q);
        auto right = lowestCommonAncestor(root->right, p, q);
        if (left && right) 
            return root;
        
        if (left) 
            return left;
        
        return right;
    
;